Exercise Solution

 

Questions

 

1. Conservation of Momentum and Closed Systems

 

In order to have conservation of momentum, it is necessary for the system to be closed. A closed system does not have any forces acting on it from the outside. A thrown ball, for example, has the force of gravity and air friction acting on it. It is not, therefore, a closed system. In order for momentum to be conserved we would have to include additional items in the system, not just the ball.

 

7. Momentum and Impulse - Advantage of "Crumple Zones" in cars

 

These zones lower the average force required for the change in momentum  incurred during the collision. They are, because of the lowered force, safer for the driver.

 

Problems

 

1.  The Definition of Momentum

 

p = mv = (0.028 kg)x(8.4 m/s) = 0.24 kg m/s

 

2.  Conservation of Momentum

 

We need to solve for the change in v so the task is to find the correct expressions to use. 

Note that this is vastly different from rote memorization or look up of equations and "blindly plugging in some numbers".

 

∆p = F∆t      and ∆p = m∆v     therefore

 

F∆t  = m∆v    gives     ∆v = F∆t/m = (25 N)x(20 s)/65 kg = -7.7 m/s which means that the skier loses 7.7 m/s of his speed.

 

15.  Collisions and Impulse

 

Recall that impulse is the change in momentum. We choose the direction of travel of the ball to be the positive direction.

 

a. Impulse imparted to the golf ball

 

∆p = m∆v = (4.5 x 10^-2 kg)x(45 m/s - 0) = 2.0 kg m/s

 

b. Average force exerted on the ball by the golf club

 

This is the impulse divided by the interaction time

 

F_average = ∆p/∆t = 2.0 kg m/s/3.5 x 10^-3 s = 5.8 x 10² N

 

17.  Collisions and Impulse - Symmetrical Problem, No Vertical Impulse (since no ∆p)

 

We know that the impulse given to the ball is just its change in momentum. We also know, from the symmetry of the problem, that the vertical momentum of the ball does not change. No change in momementum, no impulse in the vertical direction.

 

We choose the direction away from the ball as the positive direction for momentum  perp to the wall.

 

∆pperp = mv_{perp final} - mv_{perp initial} = m(v sin 45⁰ - - v sin 45⁰) = 2mvsin 45⁰

 

= (6.0 x 10^-2 km)x(25 m/s) sin 45⁰ = 2.1 kg m/s moving to the left

 

22.  Elastic Collisions, Conservation of Momentum (no outside forces)

 

In the following, A is the 0.440 kg ball and B is the 0.220 kg ball.

 

Given: v_a = 3.30 m/s     v_b = 0

 

Equation 7-7 can be used to obtain a relationship between the 2 velocities

 

v_a - v_b = -(v_a' - v_b') gives v_b' = v_a + v_a'

 

We know that momentum is conserved so, substituting the above into the momentum equation

 

(m_av_a + m_bv_b)_initial = (m_av_a + m_bv_b)_final

 

gives

 

v_afinal = 1.10 m/s east

 

vb_final = 4.40 m/s east

 

46.  Center of Mass - extended to 3 particles

 

This is just a use of equation 7-9a, extended to 3 particles

 

x _cm = (m_ax_a + m_bx_b + m_cx_c)/(m_a + m_b + m_c) = 0.44 m